# Will the limiting react always be the substance having the smallest mass?

Dec 23, 2016

Yes.

#### Explanation:

Yes. It's called the limiting reactant because it gets used up first in a chemical reaction. This results in the smallest amount of reactant in a chemical equation.

Excess is the opposite, having the largest amount.

In a reaction where there is only one product or one reactant, limiting reactants and excess do not "exist".

Hope this helps :)

Dec 23, 2016

Not necessarily.

#### Explanation:

The limiting reagent will be that with the lower quantity of moles .

When we determine the limiting reagent, we first balance the chemical equation and convert all quantities of concern to moles. Then, we use stoichiometry to determine how much product could be produced by each reactant. That which gives the lower number of moles of product is determined to be the limiting reactant. You can also simply compare the amount of moles of each reactant available. That which is present in the lower number of moles is the limiting reactant.

We cannot use the mass of each reagent present to make this determination because different elements have different molar masses, given in grams per mole. For example, if you have $120$ grams of tin (Sn) and only $60$ grams of lithium (Li), you might be inclined to think that because you have a lower mass quantity of lithium, it must be the limiting reactant. However, the molar mass of lithium is $\approx 7$ g/mol, and the molar mass of tin is $\approx 119$ g/mol. This means that you have $\approx 8.6$ moles of lithium and only $\approx 1$ mole of tin.

Here is an example of calculating the limiting reagent in the reaction of ${C}_{2} {H}_{3} B {r}_{3}$ ($76.4 g$) with ${O}_{2}$ ($49.1 g$).

Balanced chemical equation:

$4 {C}_{2} {H}_{3} B {r}_{3} + 11 {O}_{2} \to 8 C {O}_{2} + 6 {H}_{2} O + 6 B {r}_{2}$

Convert mass quantities to moles.

${C}_{2} {H}_{3} B {r}_{3}$: $266.72$ g/mol

${O}_{2}$: $32$ g/mol

$76.4 \cancel{g} {C}_{2} {H}_{3} B {r}_{3} \times \frac{1 m o l}{266.72 \cancel{g}} = 0.286$ $m o l$ ${C}_{2} {H}_{3} B {r}_{3}$

$49.1 \cancel{g} {O}_{2} \times \frac{1 m o l}{32 \cancel{g}} = 1.53$ $m o l$ ${O}_{2}$

Therefore, ${C}_{2} {H}_{3} B {r}_{3}$ is the limiting reagent, as it contributes the lower number of moles in the reaction. Notice that we began with a higher mass quantity of ${C}_{2} {H}_{3} B {r}_{3}$.

You could also calculate the moles of $C {O}_{2}$ produced by each reactant and compare using stoichiometry. In that method, whichever produces the lower amount of $C {O}_{2}$ is the limiting reactant (still ${C}_{2} {H}_{3} B {r}_{3}$).